Proving Symmetric Difference Properties of Sets

[evinda]

Hello! (Wave)

Given the following definition:

If $A,B$ are sets, we define the set $A \triangle B =(A \setminus B) \cup (B \setminus A)$ and we call it symmetric difference of $A,B$.

I have to prove the following sentences:

1. $A \cap B= \varnothing$, then $A \triangle B=A \cup B$
   
2. $A \triangle A=\varnothing$
   
3. $A \triangle B=B \triangle A$
   
4. $(A \triangle B) \triangle C= A \triangle (B \triangle C)$
   
5. $A \cap (B \triangle C)=(A \cap B) \triangle (A \cap C)$

That's what I have tried:

1. Since $A \cap B=\varnothing$, $A \setminus B=A$ and $B \setminus A=B$.
   So, $A \triangle B=(A \setminus B) \cup (B \setminus A)=A \cup B$
   
   Can we just say it like that or do we have to prove it further?
   
2. Do I have to take a $x \in A \triangle A$, or isn't it necessary?
   
   If so, then is it like that?
   
   $x \in A \triangle A \leftrightarrow x \in (A \setminus A) \cup (A \setminus A) \leftrightarrow x \in \varnothing \cup \varnothing \leftrightarrow x \in \varnothing$, that cannot be true, therefore there is no set $x$, for which $x \in A \triangle A$, so $A \triangle A=\varnothing$.
   
3. $x \in A \triangle B \leftrightarrow x \in (A \setminus B) \cup (B \setminus A) \leftrightarrow x \in (A \setminus B) \lor x \in (B \setminus A) \leftrightarrow x \in (B \setminus A) \lor x \in (A \setminus B) \leftrightarrow x \in B \triangle A $

Also, how can I show the last two sentences?

 

[mmwave]

For the third sentence, we can use the definition of symmetric difference to show that $A \triangle B = B \triangle A$:

$A \triangle B = (A \setminus B) \cup (B \setminus A)$

$= (B \setminus A) \cup (A \setminus B)$ (commutative property of union)

$= B \triangle A$ (definition of symmetric difference)

For the fourth sentence, we can use the associative property of union to show that $(A \triangle B) \triangle C = A \triangle (B \triangle C)$:

$(A \triangle B) \triangle C = ((A \setminus B) \cup (B \setminus A)) \cup C$

$= (A \setminus B) \cup ((B \setminus A) \cup C)$ (associative property of union)

$= A \triangle (B \triangle C)$ (definition of symmetric difference)

For the last sentence, we can use the distributive property of intersection over union to show that $A \cap (B \triangle C) = (A \cap B) \triangle (A \cap C)$:

$A \cap (B \triangle C) = A \cap ((B \setminus C) \cup (C \setminus B))$

$= (A \cap (B \setminus C)) \cup (A \cap (C \setminus B))$ (distributive property)

$= ((A \cap B) \setminus C) \cup ((A \cap C) \setminus B)$ (definition of set difference)

$= ((A \cap B) \cup (A \cap C)) \setminus (B \cup C)$ (distributive property)

$= (A \cap B) \triangle (A \cap C)$ (definition of symmetric difference)

Hope this helps!